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3 years ago

suppose you are in a car driving down a road. a large truck passes your car. what does the car have a tendency to do?

Physics

2 answers:

balandron [24]3 years ago

6 0

it moves toward the truck because increased air movement between the car and the truck decreases pressure.

Hope this helped :) <3

Vsevolod [243]3 years ago

5 0

Explanation:

In the given problem, you are in a car driving down a road a large truck suddenly passes by your car. The large truck has more momentum than the car due to mass.

The low pressure area is due to the higher speed of air. If the stream flow of air increases then there will be low energy pressure.

The car moves towards the truck because of the increased air movement between the car and the truck decreases pressure.

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Answer:

c she's very clever

Explanation:

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3 years ago

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A long, rigid conductor, lying along an x axis, carries a current of 4.99 A in the negative x direction. A magnetic field is pre

Tcecarenko [31]

Answer with Explanation:

We are given that

Current in conductor=I=4.99 A (-x direction)

Magnetic field=B= $3.72\hat{i}+8.72x^2\hat{j}mT=(3.72i+8.72x^2j)\times 10^{-3}$

(1mT= $10^{-3} T$ )

x(in m) and B (in mT)

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$\vec{L}=-x\hat{i}$

$dL=-dx\hat{i}$

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$F=I(L\times B)$

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Integrating on both sides then we get

$\vec{F}=\int_{1.41}^{2.77}(4.99)(-dx\hat{i}\times (3.72\hat{i}+8.72x^2\hat{j}))\times 10^{-3}$

$\vec{F}=-\int_{1.41}^{2.77}(4.99\times 10^{-3})\cdot 8.72(x^2\hat{k})dx$ ( $i\times i=0, i\times j=k$

$\vec{F}=-(4.99\times 10^{-3}\times 8.72)[\frac{x^3\hat{k}}{3}]^{2.77}_{1.41}$

$\vec{F}=-\frac{(4.99\times 10^{-3}\cdot 8.72)}{3}((2.77)^3-(1.41)^3)\hat{k}$

$\vec{F}=-0.268 \hat{k} N$

a. x- component of force=0

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3 years ago

Two workers pull horizontally on a heavy box but one pulls twice as hard as the other. The larger pull is directed at 25.0 west

icang [17]

1) Call F1 the larger force and F1x and F1y its its x-and-y- components.respectively.

I will use the complementary angle: 90 - 25 = 65 to work with the normal convention.

=> cos(65) = F1x / F1 => F1x = - F1*cos(65) (I choose negative as the west direction)

=> sin(65) = F1y / F1 => F1y = F1*sin(65) (I choose positive the north direction)

2) Call F2 the shorter force and F2x and F2y its components

=> cos(x) = F2x / F2 => F2x = F2*cos(x)

=> sin(x) = F2y / F2=> F2y = F2*sin(x)

3) You know that:

- F1 = 2F2
- The net force in the y direction is 430 N
- The net force in the x direction is 0

a) F1x + F2x = 0

=> -F1*cos(65) + F2*sin(x) = 0

=> F1*cos(65) = F2 sin(x) => sin(x) = [F1/F2] cos(65)

Remember F1 = 2F2 => F1/F2 = 2 => sin(x) = 2 cos(65) = 0.84524

=> x = arcsin(0.84524) = 57.7

b) F1y + F2y = 430 =>

F1 sin(65) + F2*sin(57.7) = 430 =>

0.9060F1 + 0.84524F2 430

F1 = 2F2 => 0.9060*2F2 + 0.84524F2 = 430 => 1.7512F2 = 430

=> F2 = 430 / 1.7512 = 245.54 N

=> F1 = 2*245.54 =491.1N

There you have the two forces.

The angle of the shorter force is 57.7 measured from the east to the north (this is north of east), which would be 90 - 57.7 = 32.3 degrees east of north..

Then the shorter force is 245.5 N at 32.3 degrees east of north

And the larger force is 491.1 N at 25.0 degrees west of north.

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3 years ago

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